**3. Method of Solution**

To adopt an ease in making the algorithm, the following primitive variables are used to make the primitive form of the above Equations (7)–(10) along with boundary conditions:

$$\begin{aligned} u &= \mathbf{x}^{\frac{1}{2}} L(\mathbf{X}, \mathbf{Y}), \quad v = \mathbf{x}^{-\frac{1}{4}} V(\mathbf{X}, \mathbf{Y}), \ \mathbf{Y} = \mathbf{x}^{-1/4} y, \ \boldsymbol{\theta} = \overline{\boldsymbol{\theta}}(\mathbf{X}, \mathbf{Y}), \ \boldsymbol{\eta} = \overline{\boldsymbol{\eta}}(\mathbf{X}, \mathbf{Y}),\\ \mathbf{x} &= \mathbf{X} \ \boldsymbol{\prime} \ \boldsymbol{\theta} = \overline{\boldsymbol{\theta}}(\mathbf{X}, \mathbf{Y}) \ \ \boldsymbol{\prime} \ \boldsymbol{\eta} = \overline{\boldsymbol{\eta}}(\mathbf{X}, \mathbf{Y}) \ \ \boldsymbol{\prime} \ \boldsymbol{\eta} = a \boldsymbol{\sin} \mathbf{X} \end{aligned} \tag{12}$$

After substitution of the variables defined in Equation (12) into Equations (7)–(11), we ge<sup>t</sup> the following primitive system of partial differential equations:

$$X \cos XUI + \left\{ X \frac{\partial II}{\partial X} - \frac{1}{4} Y \frac{\partial II}{\partial Y} + \frac{1}{2} II + \frac{\partial V}{\partial Y} \right\} \sin X = 0 \tag{13}$$

$$X\mathcal{U}\frac{\partial \mathcal{U}}{\partial X} + \frac{1}{2}\mathcal{U}^2 + \left(V - \frac{1}{4}Y\mathcal{U}\right)\frac{\partial \mathcal{U}}{\partial Y} = \frac{\partial^2 \mathcal{U}}{\partial Y^2} - \overline{\theta}\sin X - \overline{\rho}\sin X - X^{1/2}\mathcal{M}\mathcal{U} \tag{14}$$

*Mathematics* **2020**, *8*, 2010

$$X\mathcal{U}\frac{\partial\overline{\partial}}{\partial X} + \left(V - \frac{1}{4}Y\mathcal{U}\right)\frac{\partial\overline{\partial}}{\partial Y} = \frac{1}{\text{pr}}\frac{\partial^2\overline{\partial}}{\partial Y^2} + \text{Nb}\frac{\partial\overline{\varphi}}{\partial Y}\frac{\partial\overline{\partial}}{\partial Y} + \text{Nt}\left(\frac{\partial\overline{\partial}}{\partial Y}\right)^2 + \text{Q}\overline{\partial} \tag{15}$$

$$XU\frac{\partial\overline{\boldsymbol{\sigma}}}{\partial\boldsymbol{X}} + \left(V - \frac{1}{4}YU\right)\frac{\partial\overline{\boldsymbol{\sigma}}}{\partial\boldsymbol{Y}} = \frac{1}{\operatorname{Sc}}\left(\frac{\partial^2}{\partial^2\boldsymbol{Y}} + \frac{\operatorname{Nt}}{\operatorname{Nb}}\frac{\partial^2}{\partial^2\boldsymbol{Y}}\right) \tag{16}$$

The corresponding boundary conditions are:

$$\begin{aligned} \mathcal{U} = 0 &= V, & \overline{\theta} &= 1 & \overline{\varphi} &= 1 & \text{at} & \mathcal{Y} &= 0, \\ \mathcal{U} & \to 0, & \overline{\varphi} &\to 0, & \overline{\theta} &\to 0 & \text{as} & \mathcal{Y} &\to \infty \end{aligned} \tag{17}$$
